See complete Problem 96

Similar Triangles, Incenters, Parallelogram. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Geometry Problem 96

Labels:
circle,
incenter,
parallelogram,
quadrilateral,
similarity,
triangle

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join B to O1 & O extend to P ( P on AC )(same bisector)

ReplyDeleteextend GO3 to Q ( Q on AC )

ang APB = 1/2 B + C

ang MQG = 1/2 G + C ( AB//MF => ang B = ang G )

=>

ang APB = ang MQG

=>

PB // QO2

ang DEA = ang C ( DE // AC )

=>1/2 DEA = 1/2 C

=>

O2O1 // CO

To Antonio:

ReplyDeleteThe written enunciate of problem 96 is a repetition of problem 94. The figure is all right.

Thanks Nilton. The written enunciate of problem 96 has been updated.

DeleteAlternative solution to problem 96.

ReplyDeletePoints O, O1 and B are collinear, on the bisector of ang(ABC). Points O2, G and O3 are collinear, on the bisector of ang(MGC). As triangles ABC, DBE, FGE and MGC are similar,

then OB/BC = O1B/BE = O2G/EG = O3G/GC = (O1B+O2G+O3G)/(BE+EG+GC).

But BE + EG + GC = BC, so OB = O1B + O2G + O3G, which is the same as OB – O1B = O2G + O3G. Besides OO1 = OB – O1B = O2G + O3G = O2O3.

Being ABC and BGC similar, then

ang(ABO) = ang(MGO3), and, as AB // MR, then

OO1 // O2O3.

We can see now that OO1O2O3 is a parallelogram because it has two opposite sides equal and parallel.

Same as Problem 93

ReplyDelete